Components of vectors

  • Thread starter Pirang
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  • #1
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Can anyone help me with this question? I would appreciate it.

A horse pulls a barge of mass 5000 kg along a canal using a rope 10 m long. The rope is attached to a point on the barge 2m from the bank. As the barge starts to move, the tension in the rope is 500 N. Calculate the barge's initial acceleration parallel to the bank.

Please explain hehe physics is not my strong side :D
 

Answers and Replies

  • #2
Mkay, so the basic physics principle here is the relationship betwenn pulling forces and the resulting motion. In this simple case, the two interacting areas are the forces and the resulting ACCELERATION of the barge. This physical phenomenon is called Neto'ns second law of motion, which in mahematical form states that F=ma, where F is the NET force acting on a body and a is the resulting acceleration.

The question here is figuring out what information is useful here. since he formual does not need lengths of any kind, all you have left is the mass of the barge and the tension in the rope. The tension is force and the barge is a mass. I think you got it from here dude.
 
  • #3
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Damn this was easy!
I think the question is complicated and therefore hard to understand. What is the use of saying that the point the rope is connected to is 2m away from the bank or that the rope is 10m long. WHY EVEN MENTION A CANAL?

well, thank you. This was quite simple :D
 
  • #4
Hootenanny
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DaMastaofFisix said:
Mkay, so the basic physics principle here is the relationship betwenn pulling forces and the resulting motion. In this simple case, the two interacting areas are the forces and the resulting ACCELERATION of the barge. This physical phenomenon is called Neto'ns second law of motion, which in mahematical form states that F=ma, where F is the NET force acting on a body and a is the resulting acceleration.

The question here is figuring out what information is useful here. since he formual does not need lengths of any kind, all you have left is the mass of the barge and the tension in the rope. The tension is force and the barge is a mass. I think you got it from here dude.

I'm afraid this analysis is wrong. You need to resolve the tension so that it is parallel to the canal, at the moment it is cutting it diagonally across. The question only asks for the acceleration parallel to the canal, not the net acceleration. This question requires the use of trig, that is why the lengths are given.
 
  • #5
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Now that you put it this way, I figured out that sin A = 2/10 = 11.5 degrees. Since F = 500 x cos 11.5
F = 490 N approximately
F = ma

490 = 5000 x a

490/5000 = a

a = 0.098 m/seconds squared

This look correct?
 
  • #6
Hootenanny
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Pirang said:
Now that you put it this way, I figured out that sin A = 2/10 = 11.5 degrees. Since F = 500 x cos 11.5
F = 490 N approximately
F = ma

490 = 5000 x a

490/5000 = a

a = 0.098 m/seconds squared

This look correct?

Looks good to me. :smile: Note - I sent you a PM to see this thread, I didn't want you writing up incorrect work! :wink:
 

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